10. Photoelectric Effect

Introduction

You probably are used to associating a frequency ν in Hertz (or cycles/sec) and wavelength λ in cm with electromagnetic waves.  Also, you learned the relationship λν=c in the elementary physics course where c = 3.0×(10^10) cm/sec.
	This lab is more interested in the energy associated with electromagnetic waves and classically the energy density u is related to the electric field E via u=(Subscript[ϵ, 0])(E^2).  (Subscript[ϵ, 0]) is the dielectric constant related to the proportionality constant k appearing in the Coulomb force law.  Usually one is interested in the average energy so the electric field squared is averaged over a cycle.  As a result the average energy density does not depend upon the frequency (or wavelength) of the wave.  Here you will do an experiment which contradicts this classical picture.

The Photoelectric Lab Setup

A schematic of the apparatus for this experiment appears below:

The mercury light produces several well defined colors of light (yellow, green, violet and ultraviolet).  Each of these colors of the mercury light has an associated frequency f  as indicated in the table below:

({{Color, yellow, green, violet, U . V .  1, U . V .  2}, {frequency ν \[Cross](10^14) Hz, 5.19, 5.49, 6.88, 7.41, 8.22}, {wavelength λ, \[Placeholder], \[Placeholder], \[Placeholder], \[Placeholder], \[Placeholder]}})

Homework:  Calculate the wavelength λ in cm for each of these colors.  What objects in nature are also of comparable size?

The Phototube

Basically the phototube consists of two metal plates inside a glass enclosure having a vacuum.

Photons of light can enter the tube through a glass window and travel through the openings in a metal grid (a screen like material).  Eventually the photon hits the metal plate and strikes and electron in the metal plate.  If enough energy is transferred from the photon to the electron, the electron is ripped away from an atom and leaves the surface of the plate.  This electron is called a "photoelectron" since it was produced (or ejected) by a photon.  Since the electron has kinetic energy, it may travel to the grid and as a result a current flow in the external circuit with a potential V.  As indicated in the diagram above, the electron is attracted to the positive grid.  
	However, current will flow in the external circuit even if the battery potential is reversed and the grid is negative.  This might seem surprising at first but remember the photoelectron has some kinetic energy after it is ejected from the metal plate and if it has enough kinetic energy, the photoelectron can overcome a negative potential and reach the negative grid.  In the experiment, the external voltage is adjusted just so no current flows so the potential V is a measure of the amount of kinetic energy the photoelectron has when it is ejected from the plate.  Experimentally it is found the potential V depends upon the color. and you should complete the data table below:

({{Color, yellow, green, violet, U . V .  1, U . V .  2}, {frequency ν \[Cross](10^14) Hz, 5.19, 5.49, 6.88, 7.41, 8.22}, {Stopping Voltage, 0.74, \[Placeholder], \[Placeholder], \[Placeholder], \[Placeholder]}})

What is the measured uncertainly ΔV in the voltage?  You can estimate this by trying the repeat the above measurements and seeing how much V changes for a given color of light.

The Einstein Hypothesis

Einstein (1905) produced a remarkable hypothesis to explain the above experiment.  In fact Einstein's hypothesis preceded the experiment which was performed some years later by Millikan.  Einstein supposed the photon had energy related to the frequency ν

E =  n h ν

where n=1, 2, 3, ... is an integer (depending upon the number n of photons) and h is called Planck's constant.  The value of h is determined in this experiment.  Earlier Planck introduced a similar hypothesis to explain Black Body Radiation but there Planck assumed the matter oscillators had motion with energy in quanta according to equation (1).  Einstein extended Planck's idea by assuming that light (as well as matter) comes in quanta or bundles of energy.

The Photoelectric Theory

According to Einstein, the energy  E of an incoming photon is E= h ν.  This photon is a bundle or quanta of energy and it collides with an electron in the metal.  If the energy of the photon is large enough, the electron acquires enough energy to become free of the metal surface and the electron has some kinetic energy  K.E.=(1/2)m (v^2).  Conservation of energy yields

(1/2)m (v^2)=h ν - W

where W (or work function) is the potential energy lost by the electron in overcoming the attraction of the atoms in the metal.  If the voltage V is negative (the reverse of the diagram above) and adjusted so the electron just reaches the grid, then the electron has zero energy when it reaches the grid and

(1/2)m (v^2)-eV=0     or      (1/2)m (v^2)=eV

Combining equations (2) and (3) yields

eV = h ν - W

A graph of stopping potential V versus photon frequency ν is a straight line of slope h/e.

Homework for Lab Report

Graph the voltage V (vertical axis) versus frequency ν (horizontal axis) data.  Use Method of Least Squares to get best straight line V=mν+b.  Compare the value of the slope m you get with the predicted slope h/e using known value of Planck's constant h and charge of an electron e. 
	Calculate the uncertainty in the slope δm using

δm = (Sqrt[(N/Δ)]) δV

where Δ is calculated in the Method of Least Squares, N=5 is the number of colors and δV is your estimate of the uncertainty in the voltage V.
     	Treat the charge of an electron e as known and calculate the value of Planck's constant h from the slope.  Also calculate the uncertainty in Planck's constant δh using propagation of errors.  Report your measured value of Planck's constant as (Subscript[h, best]) ± δh.  Does the accepted value of Planck's constant (Subscript[h, accepted]) lie in this range?

Why the Classical Theory of Light Fails

The classical theory of light cannot explain why the energy of the photon depends upon the color of light.  Classically, there should be no dependence of voltage on frequency of light.  Also, the voltage (charge) begins to build up almost immediately after you short the grid and plate.  Classically there should be a time delay before you get the first photoelectron at the grid since an electron in the metal would need to accumulate some energy before being ejected from the metal plate.